Problem: There is a railroad in Takahashi Kingdom. The railroad consists of N sections, numbered 1, 2, ..., N, and N+1 stations, numbered 0, 1, ..., N. Section i directly connects the stations i-1 and i. A train takes exactly A_i minutes to run through section i, regardless of direction. Each of the N sections is either single-tracked over the whole length, or double-tracked over the whole length. If B_i = 1, section i is single-tracked; if B_i = 2, section i is double-tracked. Two trains running in opposite directions can cross each other on a double-tracked section, but not on a single-tracked section. Trains can also cross each other at a station.

Snuke is creating the timetable for this railroad. In this timetable, the trains on the railroad run every K minutes, as shown in the following figure. Here, bold lines represent the positions of trains running on the railroad. (See Sample 1 for clarification.)

<image>

When creating such a timetable, find the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. It can be proved that, if there exists a timetable satisfying the conditions in this problem, this minimum sum is always an integer.

Formally, the times at which trains arrive and depart must satisfy the following:

* Each train either departs station 0 and is bound for station N, or departs station N and is bound for station 0.
* Each train takes exactly A_i minutes to run through section i. For example, if a train bound for station N departs station i-1 at time t, the train arrives at station i exactly at time t+A_i.
* Assume that a train bound for station N arrives at a station at time s, and departs the station at time t. Then, the next train bound for station N arrives at the station at time s+K, and departs the station at time t+K. Additionally, the previous train bound for station N arrives at the station at time s-K, and departs the station at time t-K. This must also be true for trains bound for station 0.
* Trains running in opposite directions must not be running on the same single-tracked section (except the stations at both ends) at the same time.

Constraints

* 1 \leq N \leq 100000
* 1 \leq K \leq 10^9
* 1 \leq A_i \leq 10^9
* A_i is an integer.
* B_i is either 1 or 2.

Input

The input is given from Standard Input in the following format:


N K
A_1 B_1
A_2 B_2
:
A_N B_N


Output

Print an integer representing the minimum sum of the amount of time required for a train to depart station 0 and reach station N, and the amount of time required for a train to depart station N and reach station 0. If it is impossible to create a timetable satisfying the conditions, print -1 instead.

Examples

Input

3 10
4 1
3 1
4 1


Output

26


Input

1 10
10 1


Output

-1


Input

6 4
1 1
1 1
1 1
1 1
1 1
1 1


Output

12


Input

20 987654321
129662684 2
162021979 1
458437539 1
319670097 2
202863355 1
112218745 1
348732033 1
323036578 1
382398703 1
55854389 1
283445191 1
151300613 1
693338042 2
191178308 2
386707193 1
204580036 1
335134457 1
122253639 1
824646518 2
902554792 2


Output

14829091348

Solution (in CPP):

Solution:
#include <iostream>
#include <sstream>
#include <iomanip>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cctype>
#include <cstring>
#include <vector>
#include <list>
#include <queue>
#include <deque>
#include <stack>
#include <map>
#include <set>
#include <algorithm>
#include <iterator>
#include <bitset>
#include <ctime>
#include <complex>

using namespace std;

#define FOR(i,a,b) for (int i = (a); i < (b); i++)
#define RFOR(i,b,a) for (int i = (b)-1; i >= (a); i--)
#define ITER(it,a) for (__typeof(a.begin()) it = a.begin(); it != a.end(); it++)
#define FILL(a,value) memset(a, value, sizeof(a))

#define SZ(a) (int)a.size()
#define ALL(a) a.begin(), a.end()
#define PB push_back
#define MP make_pair

typedef long long LL;
typedef vector<int> VI;
typedef pair<int, int> PII;

const double PI = acos(-1.0);
const int INF = 1000 * 1000 * 1000 + 7;
const LL LINF = INF * (LL) INF;

const int MAX = 2* 100 * 1000 + 47;



struct RMQ
{
    int T[MAX << 4];
    
    void push(int v, int tl, int tr)
    {
        if (!T[v] || tl == tr)return;
        T[v+v] = T[v+v+1] = T[v];
        T[v] = 0;
    }
    
    void set(int v, int tl, int tr, int l, int r, int x)
    {
        if (r < l)return ;
        push(v, tl, tr);
        if (l == tl && r == tr)
        {
            T[v] = x;
            return ;
        }
        int tm = (tl + tr) / 2;
        set(v + v, tl, tm, l, min(r, tm), x);
        set(v + v + 1, tm+1, tr, max(tm+1, l), r, x);
    }
    
    int get(int v, int tl, int tr, int pos)
    {
        if (tl == tr)return T[v];
        push(v, tl, tr);
        int tm = (tl + tr) / 2;
        if (pos <= tm)return get(v + v, tl, tm, pos);
        else return get(v+v+1, tm+1, tr, pos);
    }
};

RMQ T;
int L[MAX], R[MAX];
int A[MAX];
int n, k, sz;
LL D[MAX];



LL F(int x)
{
    int f = T.get(1, 1, sz, x);
    if (!f)return 0;
    int a = A[L[f]] - A[x];
    if (a < 0)a += k;
    return D[f] + a;
}


int main()
{
    // freopen("in.txt", "r", stdin);
    scanf("%d%d", &n, &k);
    LL s = 0;
    int prev = 0;
    FOR(i, 1, n+1)
    {
        int a, b;
        scanf("%d%d", &a, &b);
        a *= 2;
        s += a;
        if (b == 1 && a > k)
        {
            printf("-1\n");
            return 0;
        }
        a = (prev - a % k + k) % k;
        if (b == 1)L[i] = prev, R[i] = a;
        else L[i] = 0, R[i] = k-1;
        prev = a;
    }
    FOR(i, 1, n+1)
        A[i + i - 1] = L[i],
        A[i + i] = R[i];
    sort(A + 1, A + n + n + 1);
    sz = unique(A + 1, A + n + n + 1) - A - 1;
    
    FOR(i, 1, n+1)
    {
        L[i] = lower_bound(A + 1, A + sz + 1, L[i]) - A;
        R[i] = lower_bound(A + 1, A + sz + 1, R[i]) - A;
    }
    
    RFOR(i, n+1, 1)
    {
        D[i] = F(L[i]);
        if (L[i] > R[i])
        {
            T.set(1, 1, sz, R[i] + 1, L[i] - 1, i);
        }
        else
        {
            T.set(1, 1, sz, 1, L[i]-1, i);
            T.set(1, 1, sz, R[i]+1, sz, i);
        }
    }

    
    
    LL ans = D[1];
    FOR(i, 1, sz+1)
    {
     //   cout << F(i) << endl;
        ans = min(ans, F(i));
    }
    ans += s;
    printf("%lld\n", ans);
    
    
    
    return 0;
}